Question

In Young's double slit experiment, the central maximum is bright in the interference pattern obtained on a screen. What will happen when light waves emitted out of slits $S_1$ and $S_2$ have an initial phase difference of $\pi$ radian ? Justify your answers.

Hint:

When sources are completely out of phase by $\pi$, the conditions for maxima and minima completely swap. Constructive interference becomes destructive and vice-versa everywhere on the screen.

Answer 1

Concept:
• In a standard Young's Double Slit Experiment (YDSE), the sources are exactly in phase (initial phase difference = $0$). At the center of the screen, the path lengths from both slits are equal ($\Delta x = 0$).
• This zero path difference normally results in a net phase difference of zero, leading to constructive interference (a central bright maximum).
• The total phase difference $\Delta \phi$ at any point on the screen is the sum of the initial phase difference $\phi_0$ and the phase difference due to the path difference ($\frac{2\pi}{\lambda} \Delta x$).

Step 1: Analyze the new phase condition at the center
The problem states that the waves emitted from slits $S_1$ and $S_2$ now have an inherent initial phase difference, $\phi_0 = \pi$ radians. At the exact geometric center of the screen, the distance travelled by the light from both slits is perfectly identical. Therefore, the path difference at the center is still zero: $\Delta x = 0$. The phase difference created by the path length difference is: \[ \phi_{path} = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times 0 = 0 \text{ radians} \]

Step 2: Calculate the total effective phase difference
The total phase difference $\Delta \phi$ at the central point is: \[ \Delta \phi = \phi_0 + \phi_{path} \]
\[ \Delta \phi = \pi + 0 = \pi \text{ radians} \]

Step 3: Determine the nature of interference
When the total phase difference between two superimposing waves is an odd multiple of $\pi$ (e.g., $\pi, 3\pi, 5\pi$), they are completely out of phase. This condition strictly leads to destructive interference. Because destructive interference minimizes intensity, a dark spot will be formed.

Step 4: Conclusion
Instead of a central bright fringe, a central dark fringe (minimum) will be obtained on the screen. The entire interference pattern will effectively shift, swapping the positions of all maxima and minima.